Let B be Standard Brownian Motion, $B_0 =0$
Consider the SDE: $dX_t = (6 + 3X_t)dt +(2+X_t)dB_t$, for the stochastic process $X = (X_t)_{t \geqslant 0}$, where $X_0 = 1$
The coefficients to this SDE satisfy the Libschitz condition, so there is a unique strong solution $X$ to this equation. It is given by:
$X_t = Y_t(1 + 4 \int_0^t \frac{1}{Y_s}ds + 2\int_0^t \frac{1}{Y_s}dB_s)$ for $t \geqslant 0$
where, $dY_t = 3Y_tdt + Y_tdB_t$ with $Y_0 =1$
Questions:
1) Compute $\mathbb{E}[X,Y]$, where $[X,Y]$ := Quadratic variation
2) Show the following holds in law:
$X_t -Y_t(1+2\int_0^t \frac{1}{Y_s}dB_s) \stackrel{law}{=}4\int_0^t Y_sds$
My attempt:
1) $[X,Y] = \frac{1}{4}([X+Y, X+Y] - [X-Y, X-Y])$
Bearing in mind that: $Y_t = 1+ \int 3Y_sds + \int Y_sdB_s$
$[X+Y, X+Y] = \int_0^t Y_s^2ds + 4Y_t^2 \int_0^t \frac{1}{Y_s^2}ds$
$[X-Y, X-Y] = 4Y_t^2 \int_0^t \frac{1}{Y_s^2}ds$
so $\mathbb{E}[X,Y] = \frac{1}{4}\mathbb{E}\int_0^t Y_s^2ds = \frac{1}{4}\int_0^t \mathbb{E}Y_s^2ds$
$dY_t$ denotes SDE for Geometric Brownian motion so $Y_t = exp(B_t - \frac{s^2}{2})$
.....then result follows quite easily.
However, was my manipulations with the quadratic variation correct?
2) Plugging in X_t into the question yields:
$Y_t\int _0^t \frac{1}{Y_s}ds \stackrel{law}{=} \int_{0}^{t} Y_sdB_s$
If both sides were a strong solution to an SDE then by uniqueness they'd be equal in law. However I am not sure how to generate such an SDE. Any ideas?
1)The quadratic variation computation we simply need to isolate the Brownian motion terms since $[dt,dt]=[dt,dB_{t}]=0$. We let
$$Z_{t}:=1 + 4 \int_0^t \frac{1}{Y_s}ds + 2\int_0^t \frac{1}{Y_s}dB_s$$
and so $X_{t} = Y_{t} Z_{t}$ and by Itô product rule we have
$$dX_{t}=Y_{t}dZ_{t}+Z_{t}dY_{t}+[dY_{t},dZ_{t}]=Y_{t}dZ_{t}+Z_{t}dY_{t}+2dt.$$
Therefore,
$$d[X_{t},Y_{t}]=Y_{t}[dZ_{t},dY_{t}]+Z_{t}[dY_{t},dY_{t}]+2[dt,dY_{t}]$$
$$=Y_{t}2dt+Z_{t}Y_{t}^{2}dt+0=Y_{t}(2+X_{t})dt$$
and so
$$E[X_{t},Y_{t}]=\int^{t} E[Y_{s}(2+X_{s})]ds.$$
2)Here we indeed need to use that $Y_{t}$ is a geometric Brownian motion
$$Y_{t}=exp(B_{t}+\frac{5}{2}t)$$
and so $$\frac{Y_{t}}{Y_{s}}=exp(B_{t}-B_{s}+\frac{5}{2}(t-s))\stackrel{d}{=}exp(B_{t-s}+\frac{5}{2}(t-s))=Y_{t-s}.$$
Using change of variables $\int^{t}_{0}Y_{t-s}ds=\int^{t}_{0}Y_{s}ds$ we get the result.